Question

ammonia nh₃ (g), δh_f = -45.9 kj/mol reacts with oxygen to produce nitrogen and water h₂o (g), δh_f = -241.8 kj/mol according to the equation. 4nh₃ (g) + 3o₂ (g) → 2n₂ (g) + 6h₂o (g) what is the enthalpy of the reaction? use δh_rxn = σ (δh_f, products) - σ (δh_f, reactants). options: 3,572.1 kj; 1,267.2 kj; -1,267.2 kj; -3,572.1 kj

Explanation:

Step1: Identify $\Delta H_f$ for each substance

• For $\ce{NH3(g)}$, $\Delta H_f = -45.9\ \text{kJ/mol}$
• For $\ce{O2(g)}$, $\Delta H_f = 0\ \text{kJ/mol}$ (element in standard state)
• For $\ce{N2(g)}$, $\Delta H_f = 0\ \text{kJ/mol}$ (element in standard state)
• For $\ce{H2O(g)}$, $\Delta H_f = -241.8\ \text{kJ/mol}$

Step2: Calculate $\sum (\Delta H_{f, \text{products}})$

Products: $2\ce{N2(g)} + 6\ce{H2O(g)}$
$\sum (\Delta H_{f, \text{products}}) = [2 \times 0] + [6 \times (-241.8)] = 0 + (-1450.8) = -1450.8\ \text{kJ}$

Step3: Calculate $\sum (\Delta H_{f, \text{reactants}})$

Reactants: $4\ce{NH3(g)} + 3\ce{O2(g)}$
$\sum (\Delta H_{f, \text{reactants}}) = [4 \times (-45.9)] + [3 \times 0] = -183.6 + 0 = -183.6\ \text{kJ}$

Step4: Apply the formula $\Delta H_{\text{rxn}} = \sum (\Delta H_{f, \text{products}}) - \sum (\Delta H_{f, \text{reactants}})$

$\Delta H_{\text{rxn}} = (-1450.8) - (-183.6) = -1450.8 + 183.6 = -1267.2\ \text{kJ}$

Answer:

$-1,267.2\ \text{kJ}$ (the option with -1,267.2 kJ)